Suppose $L$ is a finite-dimensional Lie algebra over the field of characteristic $0$. Let $\operatorname{Rad}(L)$ denote the radical of $L$. My question is:
Does there always exist an invertible derivation of $[L, \operatorname{Rad}(L)]$ ?
Motivation. Jacobson [1] proved that if $L$ is a finite-dimensional Lie algebra over the the field of characteristic $0$, and $L$ has an invertible derivation, then $L$ is nilpotent.
If the answer to my question is "yes", then together with Jacobson's result we would immediately get that $[L, \operatorname{Rad}(L)]$ is nilpotent. An independent (though somewhat long-winded) proof of the latter fact is given in Lemma C.20 in page 484 of Fulton and Harris [2].
[1] N. Jacobson, A note on automorphisms and derivations of Lie algebras, Proc. of AMS. 6 (1955), 281 − 283.
[2] W. Fulton, J. Harris, Representation theory: A first course. Graduate Texts in Mathematics. Springer-Verlag, New York, 1991.
The answer is no. Suppose for example that $L$ is solvable, so that $[L,Rad(L)]=[L,L]$ is a nilpotent Lie algebra. As it is well-known, there are plenty of nilpotent Lie algebras of dimension $n>6$ which only have nilpotent derivations (they are called characteristically nilpotent Lie algebras). It is possible to chose a solvable Lie algebra $L$ such that $[L,L]$ is a characteristically nilpotent Lie algebra. So there exists no invertible derivation of $[L,Rad(L)]=[L,L]$ then.